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anonymous
 5 years ago
For a geometric sequence, a1 = 2 r=2 and an= 64. Find n and Sn The n's are subscript, but I don't know how to type them like that on here.
anonymous
 5 years ago
For a geometric sequence, a1 = 2 r=2 and an= 64. Find n and Sn The n's are subscript, but I don't know how to type them like that on here.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A geometric sequence is much like a geometric series in that the formula is as follows: \[a_{n}=ar^{n}\] when n starts at 0 or\[a_{n}=ar^{n1}\] when n starts at 1. Since you are given \[a_{1} = 2\] Then we will use the \[a_{n}=ar^{n1}\]formula because that starts at n=1, so n can be deduced by plugging the values given in the formula, i.e.\[a_{1}=2=ar^{11}=ar^{0}=a\]\[a_{n}=64=2*r^{n1}\]\[64/2=2^{n1}=2^{n}2^{1}\]\[64/2=2^{n}/2\]\[64=2^{n}\]\[\log_{2}(64)=n=6\]So this saya that the last value in the sequence is 64 and that term is n=6. Sn is in the next post

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From above, \[s_{n}=ar^{11}+ar^{21}+ar^{31}+...+ar^{n1}\] \[s_{n}=a+ar+ar^{2}+...+ar^{n1}\] Now if we multiply r on both sides of this equation, we get \[rs_{n}=ar+ar^{2}+ar^{3}+...+ar^{n1}r\] The last term is \[ar^{n1}r=ar^{n1+1}=ar^{n}\] Subtract rSn from Sn \[s_{n}rs_{n}=a+ar+ar^{2}+...+ar^{n1} \] \[ (ar+ar^{2}+ar^{3}+...+ar^{n})\] Notice that all the terms cancel each other except the first and the last. \[s_{n}rs_{n}=aar^{n}\]\[s_{n}(1r)=a(1r^{n})\]\[s_{n}=a(1r^{n})/(1r)=2(12^{n})/(1)=2(12^{n})\]Knowing that a1=a
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